A=p(1+i/m)^mn
A=3,500×(1+0.0229÷12)^(12×6)
A=4,014.98
Answer: 29.28 years
Explanation:
From Kepler's third law the square of orbital period of revolving celestial body is proportional to the cube of semi -major axis from the body it is revolving about.
P² =A³
Where, P is the orbital period in years and A is the semi-major axis in AU (Astronomical units)
It is given that, For Saturn, A = 9.5 AU. We need to find P
⇒P² = (9.5 AU)³
⇒P² = 857.38
⇒P = 29.28 years
Thus, the orbital period of Jupiter is 29.28 years around the Sun.
The factored forms of that is (x + 6) (x - 7).
If you have any questions then please leave a comment. Good luck!
This is combinations
exg. lets say you have 2 hats and 1 pair of shoes, the answe ris 2 options, if 2 hats and 2 pairs of shoes, the answer is 4 because 2 times 2=4
3 options for sandwitch
3 soup
4 salad
2 drink
we multiply all together
3 times 3 times 4 times 2=72 options
